Lesson video

Hello, Mr Robson here, welcome to Maths.

Today, we're estimating journeys from linear speed-time graphs.

Well, that sounds like a pretty fun bit of mathematics.

Should we take a look? Our learning outcome is that we can deduce that the area under a speed-time graph gives the distance travelled and deduce how this may be calculated.

Keywords, compound shape.

A compound shape is a shape created using two or more basic shapes.

Two parts to our learning today.

We're going to begin by looking at the distance from a speed-time graph.

Izzy and Andeep are discussing this speed-time graph of a car journey.

Izzy asks, "What does this graph tell us?" Andeep says, "The car travelled at 50 miles per hour for 10 hours." Izzy says, "Can you see what else it tells us?" "Not yet," says Andeep.

Izzy continues to probe.

"How far has the car travelled after one hour?" That'll be 1 lot of 50, 50 miles.

That's time multiplied by speed giving us the distance.

"How about 2 hours?" Says Izzy.

"Well, that would be 2 times 50, 100 miles," says Andeep.

"Three?" "3 times 50, 150 miles," says Andeep.

"Have you noticed anything yet? Challenges Izzy.

I'll ask you the same question.

Have you noticed what's going on here? Pause, have a think.

Welcome back, I wonder what you've noticed.

Did you notice what Andeep's noticed? This is a glorious revelation from Andeep.

"The area bound by the graph and the horizontal axis tells us the distance travelled!" That's true, on a speed-time graph.

Andeep looks at this example.

"5 lots of 50 makes 250 miles, after five hours the car has travelled 250 miles." After 10 hours, the car has travelled 10 lots of 50, 500 miles." "Well done, Andeep," says Izzy.

I second that praise.

Quick check you've got this.

How far has the car in this speed-time graph travelled after 4 hours? Four options, pause, and take your pick.

Welcome back, hopefully, you said C, 260 miles.

Where does that come from? It's this area here.

4 hours at 65 miles an hour, will indeed be 260 miles.

We won't always see a horizontal line representing speed on our speed-time graph.

For a diagonal line, the fact remains, the area bound by the line and the horizontal axis represents the distance travelled.

If we want to know how far this vehicle has travelled after four seconds, we need to work out the area of that triangle, that's half of the base time's height.

So for this one, that's 4 multiplied by 10 over 2, that's gonna be 20 metres.

Comparing these two graphs shows us why area equals distance, for a diagonal line or a horizontal line.

A rate of constant acceleration from 0 to 10 metres per second over 4 seconds would cover the same distance as a constant speed of 5 metres per second over 4 seconds.

If you think about that graph on the right, it has the same average speed as the graph on the left.

Quick check you've got this so far.

Here are three graphs of people running.

Which person ran the greatest distance? Pause, and work this one out.

Welcome back.

Hopefully, you found the area under that diagonal line from 0 to 5 seconds, that'll be 5 by 10 over 2, 25 metres.

For B, the area bound by that diagonal line and the horizontal axis is the triangle, which is 6 by 10 halved, that's 30 metres, and for C, 8 by 8 over 2 gives us 32 metres.

So it was C who travelled the furthest.

Izzy and Andeep are discussing a different speed-time graph.

Andeep says, "So the area of this triangle will tell us the distance travelled.

4 by 2 over 2, that's 4." Andeep is not correct.

Explain how you know immediately that he is incorrect.

Pause, have a conversation with the person next to you or a good thing to yourself, see you in a moment.

Welcome back.

Izzy explains this one to Andeep.

"You know the distance cannot be 4 metres because someone running just one second at 6 metres per second will cover 6 metres.

The answer has to be greater than 4." Nice explanation, Izzy.

Andeep has realised his mistake.

He reflects and says, "Of course.

I need the whole area bound by the graph and the horizontal axis! So this would be the triangle plus the rectangle.

It's 28 metres." It is, but Izzy he as a challenge.

"Yes, but you could work this one out more efficiently." What do you think Izzy means? Can you see a more efficient way to work this area out? Pause, and have a think.

Welcome back, I wonder what you said.

"It's a trapezium," said Andeep.

Can you see the trapezium? To find the area of a trapezium we do half of the upper base plus the lower base, multiplied by the height, in this case, that would be 1/2 of (6 + 8) multiplied by 4, giving us the same answer, 28 metres.

A different perspective might make this trapezium easier to see.

I've rotated that graph by 90 degrees.

Can you now see where I got the lengths 6, 8, and 4 from to substitute into the area for the trapezium? "Well done, Andeep!" That is more efficient," says Izzy.

Quick check you've got this.

Which of these accurately calculates the distance travelled on this speed-time graph? I'd like to know the distance travelled after 4 seconds, there's four options there, pause, and take your pick.

Welcome back, let's see how we did.

Option A was not an option.

Option B was a good one.

4 by 4 plus 5 by 4 over 2 is the area of the rectangle plus the area of the triangle, gives us 26 metres, this works, it's accurate, but it's not the most efficient way.

C did not work.

D worked beautifully.

Treating the shape as one trapezium is the most efficient way to calculate the distance travelled on this speed-time graph.

Practise time now.

Question 1, I'd like to order these speed-time graphs from shortest to longest in terms of distance travelled.

Pause, and do that now.

For Question 2, I'd like you to order these graphs of three snippets of car journeys from shortest to longest in terms of distance travelled.

Pause, and do this now Question 3.

This speed-time graph has no scale on the vertical axis.

Izzy says, "I've been told that the distance travelled after 4 seconds is 20 metres.

This is a graph showing direct proportion so I know the distance travelled after 8 seconds will be 40 metres." Show that this is not the case and write at least one sentence to explain Izzy's misunderstanding.

Pause, and do this now.

Feedback time now, let's see how we go on with Question 1.

Hopefully, for A, you found the area travelled to be 18 metres.

For B, the area travelled was 32 metres, and for C, the area travelled was 28 metres.

Therefore, we should have said that A's distance is less than C's distance is less than B's distance.

Question 2, we should have found an area of 95 underneath that graph, it's a distance of 95 metres, most efficiently done with a trapezium.

I do hope you calculated it as one trapezium.

For B, the area of that trapezium is 78, so that's 78 metres travelled, for C, we've got 84 metres travelled, so we should have said B is less than C is less than A, that was the right order.

Question 3 was a tricky one because you want Izzy to be right.

It is a graph of direct proportion.

So what's going on here? Why is Izzy not right? Firstly, we're going to show that this is not the case.

The distance after 4 seconds we were told was 20 metres.

Now if that's the case, we know that reading from the vertical axis at that point must be 10.

That point has to represent a speed of 10 metres per second because 10 multiplied by 4 over 2, will give us that distance of 20 metres.

If that point on the vertical axis is 10, and that point at the top of the axis is 20, so when we come to look at how far we've travelled after 8 seconds now, 8 multiplied by 20 over 2 is 80 metres.

Now that we've shown Izzy that we couldn't just double the 20 metres to 40, we've got to explain why this is not the case.

You might have written, "This is a graph of direct proportion, but the variables that are in direct proportion to one another are 'Speed' and 'Time;' the speed after 8 seconds is indeed double the speed after 4 seconds.

But distance is the area bound by the graph and the horizontal axis, therefore it is not in direct proportion to speed nor time." Onto the second half of our learning now, where we're going to look at more complex speed-time graphs.

Speed-time graphs are not always straight line graphs.

These two lines show a vehicle accelerating then decelerating.

This graph does not show a constant increase of speed, but we still calculate the distance travelled by calculating the area bound by the graph and the horizontal axis.

One option for doing this is to consider this as two triangles, a compound shape, if you will.

Triangle one, up to 8 seconds, has the dimensions 8 by 18, half that, and then after 8 seconds to 20 seconds, well, that's 12 by 18, half that.

72 plus 108 gives us an area of 180, so we've travelled 180 metres.

That's one option, but it's not the most efficient.

The most efficient will be to treat this as one triangle.

We've got a base of 20, a height of 18, 20 by 18 over 2, well, that's 180 metres.

The exact same answer, just a lot more efficient.

We mathematicians, we love efficiency.

This speed-time graph has three parts of its journey acceleration, then a constant speed, and then deceleration.

How can we calculate the distance travelled? I'd like you to try and devise at least three different ways to calculate the distance travelled in these 20 seconds.

Once you've had to go at least three different ways to break that area up into different areas, which one do you prefer, and why? Pause, have a good play with this graph.

Welcome back.

Hopefully, you found multiple ways to tackle the same problem, and found the same area, therefore distance travelled each time.

I'm gonna show you two ways it could have been done.

This one's quite neat, to consider it as three shapes, a triangle up to 4 seconds from 4 to 12 seconds, that's a rectangle.

And then beyond 12 seconds, it's another triangle, that'll give you those three areas, which when we sum them together gives us an area of 196, therefore, the distance travelled is 196 metres.

That's an accurate way of doing it, but there is a more efficient way.

Did you spot this? It's one trapezium.

We could have done 1/2 multiplied by (8 + 20) multiplied by 14, and got to 196 metres a lot quicker.

Quick check you've got this.

I'd you like to calculate the distance travelled by this vehicle using two different methods, and comment on the difference in efficiency between your two methods.

Pause, and do this now.

Welcome back, let's see how we did.

One option was to consider this as three shapes, a triangle, a rectangle, and a triangle, that would've given you areas of 40, 60, and 10, giving you total area of 110 there.

therefore, there's 110 metres travelled.

Option two, and most efficient option, was to consider it as one shape, a trapezium, whereby we would've done half multiplied by brackets 6 plus 16 multiplied by 10, and we get to 110 metres, a lot more efficiently.

Treating this as one shape is the most efficient way to calculate this one.

For some speed-time graphs, it is best to calculate the distance travelled by treating the area as a compound shape, rather than one shape.

This is a quadrilateral, but it's not one with a simple formula for area that we could use.

It's better to consider this one as two shapes.

By putting a vertical line there, we've got a triangle and a trapezium.

The area of triangle is 8 by 8 over 2.

The area of the trapezium is a half of brackets 8 plus 20 multiplied by 4.

The sum of those two areas together, we have a total area of 88, therefore, this distance travelled is 88 metres.

There are still considerations to be made over efficiency.

Izzy looks at this example and says, "I see this area as four shapes." Andeep looks at it and says, "I see it as two." Whose method do you think will be more efficient? Pause, have a conversation with the person next to you, or a good think to yourself.

See you in a moment.

Welcome back, I wonder what you said.

Did you go for Izzy's method, Andeep's method? Izzy had four shapes, a rectangle, a triangle, a rectangle, and a triangle.

She had to work out those four areas, and sum them together, she got 172 metres, that's right, that's a lot of maths there.

Andeep's was a little more efficient.

Just two shapes, two trapezium, which can be worked out a lot quicker.

"You win," says Izzy.

I think I agree.

Quick check you can do that.

I'd like you to efficiently find the distanced travelled in this speed-time graph.

Pause, and give this one a go now.

Welcome back, let's see how we did.

The most efficient way will be to break it down into those three shapes, a trapezium, a rectangle, and a trapezium.

The first trapezium is half of bracket 4 plus 6 times 2.

The rectangle 5 by 6, and the last trapezium half of brackets 6 plus 10 multiply by 3.

Add those three areas together, we get an area route 64, therefore, on a speed-time graph, that's 64 metres travelled.

Practise time now.

For question 1, I'd like you to efficiently calculate the distance travelled by a car as modelled in this speed-time graph.

You can pause and do this now.

For question 2.

Izzy calculates the distance travelled by a racing car in the first 20 seconds of a race, as modelled by this speed-distance graph.

Izzy says, "It's 73 metres in total." Write at least two bits of feedback, giving suggestions to Izzy on how she could improve her work.

I'd like at least two sentences for this one.

Pause, and do this now.

Feedback time, let's see how we got on.

Question 1, we were calculating the distance travelled by a car as modelled in this speed-time graph, and most importantly, doing it efficiently.

Hopefully, you broke it into three shapes, a trapezium, a rectangle, and a trapezium.

Those areas would have been 32, and 66, and 57 giving us a sum of 155.

In the context of a speed-time graph, that area becomes 155 metres travelled.

For question 2, Izzy was calculating the distance travelled by a racing car in the first 20 seconds.

You can see on the graph how she did that, and she declared it's 73 metres in total, and I asked you for at least two pieces of feedback for Izzy.

You might have written: "Read the scale on the axis rather than counting squares when calculating distance." You can see from that rectangle there.

Izzy's counted the squares here, there's 30 squares, but that doesn't mean that's 30 metres travelled.

That area there shows 6 seconds at 40 metres per second, which is 240 metres in distance, it's not 30.

You might also have written: "Consider efficiency.

You could have broken this area into fewer shapes," by inserting vertical lines like so, you break it down into an efficient set of shapes, the average which we could calculate.

You might also have written: "Consider your answer in context.

Dina Asher-Smith can run 100 metres in just over 10 seconds, so a racing car travelling 73 metres in 20 seconds feels too small a distance to be the right answer." Sadly, we've reached the of the lesson now, but we have learned, the area under a speed-time graph is the distance travelled, and we can deduce why this is so.

We also know that the distance travelled can be calculated in a variety of ways from a speed-time graph, but some approaches are more efficient, and you know we mathematicians love efficiency.

Well, I've enjoyed this lesson.

I hope you have too, and I look forward to seeing you again soon for more mathematics.

Goodbye, for now.